Solve the system of linear equations using matrices. x+y+z=3 2x+3y+2z=7 3x-4y+z=4

Solve the system of linear equations using matrices.
x+y+z=3
2x+3y+2z=7
3x-4y+z=4
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Step 1
Given equation
x+y+z=3
2x+3y+2z=7
3x-4y+z=4
in matrix form
$A=\left[\begin{array}{ccc}1& 1& 1\\ 2& 3& 2\\ 3& -4& 1\end{array}\right]$
$B=\left[\begin{array}{c}3\\ 7\\ 4\end{array}\right]$
$X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$
we know that $A\cdot X=B$
So, $X={A}^{-1}\cdot B$
Step 2
Now,
$X={A}^{-1}\cdot B$
We have to find ${A}^{-1}$
Adjoin the identity matrix onto the right of the original matrix, so that you have A on the left side and the identity matrix on the right side. $\left[\begin{array}{ccc}1& 1& 1\\ 2& 3& 2\\ 3& -4& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
now we will apply the elementary row method ${R}_{2}-2{R}_{1}\to {R}_{2}$ (multiply 1 row by 2 and subtract it from 2 row), ${R}_{3}-3{R}_{1}\to {R}_{3}$ (multiply 1 row by 3 and subtract it from 3 row)
$\left[\begin{array}{ccc}1& 1& 1\\ 2-2\cdot 1& 3-2\cdot 1& 2-2\cdot 1\\ 3-3\cdot 1& -4-3\cdot 1& 1-3\cdot 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0-2\cdot 1& 1-2\cdot 0& 0-2\cdot 0\\ 0-3\cdot 1& 0-3\cdot 0& 1-3\cdot 0\end{array}\right]\phantom{\rule{0ex}{0ex}}\left[\begin{array}{ccc}1& 1& 1\\ 0& 1& 0\\ 0& -7& -2\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ -2& 1& 0\\ -3& 0& 1\end{array}\right]$
${R}_{1}-1{R}_{2}\to {R}_{1}$ (multiply 2 row by 1 and subtract it from 1 row), ${R}_{3}+7{R}_{2}\to {R}_{3}$ (multiply 2 row by 7 and add it to 3 row)
$\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 0& 0& -2\end{array}\right]=\left[\begin{array}{ccc}3& -1& 0\\ -2& 1& 0\\ -17& 7& 1\end{array}\right]$
$\frac{{R}_{3}}{-2}\to {R}_{3}$ (divide the 3 row by -2)
$\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}3& -1& 0\\ -2& 1& 0\\ 8.5& -3.5& -0.5\end{array}\right]$
${R}_{1}-1{R}_{3}\to {R}_{1}$ (multiply 3 row by 1 and subtract it from 1 row)
$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}-5.5& 2.5& 0.5\\ -2& 1& 0\\ 8.5& -3.5& -0.5\end{array}\right]$
${A}^{-1}=\left[\begin{array}{ccc}-5.5& 2.5& 0.5\\ -2& 1& 0\\ 8.5& -3.5& -0.5\end{array}\right]$
Step 3
Now,
$X={A}^{-1}\cdot B$
Jeffrey Jordon